2007 USAMO Problems/Problem 5: Difference between revisions

Revision as of 19:25, 15 October 2007

Problem

Prove that for every nonnegative integer $n$ , the number $7^{7^n}+1$ is the product of at least $\displaystyle 2n+3$ (not necessarily distinct) primes.

Solution

Solution 1

We proceed by induction.

Let ${a_{n}}$ be $7^{7^{n}}+1$ . The result holds for ${n=0}$ because ${a_0 = 2^3}$ is the product of $3$ primes.

Now we assume the result holds for ${n}$ . Note that ${a_{n}}$ satisfies the recursion

${a_{n+1}= (a_{n}-1)^{7}+1} = a_{n}\left(a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}\right)$ .

Since ${a_n - 1}$ is an odd power of ${7}$ , ${7(a_n-1)}$ is a perfect square. Therefore ${a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}}$ is a difference of squares and thus composite, i.e. it is divisible by ${2}$ primes. By assumption, ${a_n}$ is divisible by ${2n + 3}$ primes. Thus ${a_{n+1}}$ is divisible by ${2+ (2n + 3) = 2(n+1) + 3}$ primes as desired.

Solution 2

Notice that $7^{{7}^{k+1}}+1=(7+1) \frac{7^{7}+1}{7+1} \cdot \frac{7^{7^2} + 1}{7^{7^1} + 1} \cdots \frac{7^{7^{k+1}}+1}{7^{7^{k}}+1}$ . Therefore it suffices to show that $\frac{7^{7^{k+1}}+1}{7^{7^{k}}+1}$ is composite.

Let $x=7^{7^{k}}$ . The expression becomes

$\frac{7^{7^{k+1}}+1}{7^{7^{k}}+1} = \frac{x^7 + 1}{x + 1}$

which is the shortened form of the geometric series $\displaystyle x^{6}-x^{5}+x^{4}-x^3 + x^2 - x+1$ . This can be factored as $\displaystyle (x^{3}+3x^{2}+3x+1)^{2}-7x(x^{2}+x+1)^{2}$

Since $x$ is an odd power of $7$ , $\displaystyle 7x$ is a perfect square, and so we can factor this by difference of squares. Therefore, it is composite.

@@ Line 9: / Line 9: @@
 We proceed by induction.
-Let <math>\displaystyle{a_{n}}</math> be <math>7^{7^{n}}+1</math>. The result holds for <math>\displaystyle{n=0}</math> because <math>\displaystyle{a_0 = 2^3}</math> is the product of <math>\displaystyle{3}</math> primes.
+Let <math>{a_{n}}</math> be <math>7^{7^{n}}+1</math>. The result holds for <math>{n=0}</math> because <math>{a_0 = 2^3}</math> is the product of <math>3</math> primes.
-Now we assume the result holds for <math>\displaystyle{n}</math>. Note that <math>\displaystyle{a_{n}}</math> satisfies the [[recursion]]
+Now we assume the result holds for <math>{n}</math>. Note that <math>{a_{n}}</math> satisfies the [[recursion]]
-<div style="text-align:center;"><math>\displaystyle{a_{n+1}= (a_{n}-1)^{7}+1} = a_{n}\left(a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}\right)</math>.</div>
+<div style="text-align:center;"><math>{a_{n+1}= (a_{n}-1)^{7}+1} = a_{n}\left(a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}\right)</math>.</div>
-Since <math>\displaystyle{a_n - 1}</math> is an odd power of <math>\displaystyle{7}</math>, <math>\displaystyle{7(a_n-1)}</math> is a perfect square. Therefore <math>\displaystyle{a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}}</math> is a difference of squares and thus [[composite]], i.e. it is divisible by <math>\displaystyle{2}</math> primes. By assumption, <math>\displaystyle{a_n}</math> is divisible by <math>\displaystyle{2n + 3}</math> primes. Thus <math>\displaystyle{a_{n+1}}</math> is divisible by <math>\displaystyle{2+ (2n + 3) = 2(n+1) + 3}</math> primes as desired.
+Since <math>{a_n - 1}</math> is an odd power of <math>{7}</math>, <math>{7(a_n-1)}</math> is a perfect square. Therefore <math>{a_{n}^{6}-7(a_{n}-1)(a_{n}^{2}-a_{n}+1)^{2}}</math> is a difference of squares and thus [[composite]], i.e. it is divisible by <math>{2}</math> primes. By assumption, <math>{a_n}</math> is divisible by <math>{2n + 3}</math> primes. Thus <math>{a_{n+1}}</math> is divisible by <math>{2+ (2n + 3) = 2(n+1) + 3}</math> primes as desired.
 === Solution 2 ===

2007 USAMO (Problems • Resources)
Preceded by Problem 4	Followed by Problem 6
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